3.15.91 \(\int (A+B x) (d+e x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx\)

Optimal. Leaf size=164 \[ \frac {x^2 \sqrt {a^2+2 a b x+b^2 x^2} (a A e+a B d+A b d)}{2 (a+b x)}+\frac {x^3 \sqrt {a^2+2 a b x+b^2 x^2} (a B e+A b e+b B d)}{3 (a+b x)}+\frac {a A d x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {b B e x^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (a+b x)} \]

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Rubi [A]  time = 0.08, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {770, 77} \begin {gather*} \frac {x^3 \sqrt {a^2+2 a b x+b^2 x^2} (a B e+A b e+b B d)}{3 (a+b x)}+\frac {x^2 \sqrt {a^2+2 a b x+b^2 x^2} (a A e+a B d+A b d)}{2 (a+b x)}+\frac {a A d x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {b B e x^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)*(d + e*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

(a*A*d*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a + b*x) + ((A*b*d + a*B*d + a*A*e)*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]
)/(2*(a + b*x)) + ((b*B*d + A*b*e + a*B*e)*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*(a + b*x)) + (b*B*e*x^4*Sqrt[
a^2 + 2*a*b*x + b^2*x^2])/(4*(a + b*x))

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int (A+B x) (d+e x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (a b+b^2 x\right ) (A+B x) (d+e x) \, dx}{a b+b^2 x}\\ &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (a A b d+b (A b d+a B d+a A e) x+b (b B d+A b e+a B e) x^2+b^2 B e x^3\right ) \, dx}{a b+b^2 x}\\ &=\frac {a A d x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {(A b d+a B d+a A e) x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac {(b B d+A b e+a B e) x^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac {b B e x^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (a+b x)}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 74, normalized size = 0.45 \begin {gather*} \frac {x \sqrt {(a+b x)^2} (2 a (3 A (2 d+e x)+B x (3 d+2 e x))+b x (A (6 d+4 e x)+B x (4 d+3 e x)))}{12 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)*(d + e*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

(x*Sqrt[(a + b*x)^2]*(2*a*(3*A*(2*d + e*x) + B*x*(3*d + 2*e*x)) + b*x*(B*x*(4*d + 3*e*x) + A*(6*d + 4*e*x))))/
(12*(a + b*x))

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IntegrateAlgebraic [F]  time = 1.12, size = 0, normalized size = 0.00 \begin {gather*} \int (A+B x) (d+e x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(A + B*x)*(d + e*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

Defer[IntegrateAlgebraic][(A + B*x)*(d + e*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2], x]

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fricas [A]  time = 0.41, size = 52, normalized size = 0.32 \begin {gather*} \frac {1}{4} \, B b e x^{4} + A a d x + \frac {1}{3} \, {\left (B b d + {\left (B a + A b\right )} e\right )} x^{3} + \frac {1}{2} \, {\left (A a e + {\left (B a + A b\right )} d\right )} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)*((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

1/4*B*b*e*x^4 + A*a*d*x + 1/3*(B*b*d + (B*a + A*b)*e)*x^3 + 1/2*(A*a*e + (B*a + A*b)*d)*x^2

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giac [A]  time = 0.16, size = 114, normalized size = 0.70 \begin {gather*} \frac {1}{4} \, B b x^{4} e \mathrm {sgn}\left (b x + a\right ) + \frac {1}{3} \, B b d x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{3} \, B a x^{3} e \mathrm {sgn}\left (b x + a\right ) + \frac {1}{3} \, A b x^{3} e \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, B a d x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, A b d x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, A a x^{2} e \mathrm {sgn}\left (b x + a\right ) + A a d x \mathrm {sgn}\left (b x + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)*((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

1/4*B*b*x^4*e*sgn(b*x + a) + 1/3*B*b*d*x^3*sgn(b*x + a) + 1/3*B*a*x^3*e*sgn(b*x + a) + 1/3*A*b*x^3*e*sgn(b*x +
a) + 1/2*B*a*d*x^2*sgn(b*x + a) + 1/2*A*b*d*x^2*sgn(b*x + a) + 1/2*A*a*x^2*e*sgn(b*x + a) + A*a*d*x*sgn(b*x +
a)

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maple [A]  time = 0.05, size = 76, normalized size = 0.46 \begin {gather*} \frac {\left (3 B b e \,x^{3}+4 x^{2} A b e +4 x^{2} a B e +4 b B d \,x^{2}+6 x a A e +6 x A b d +6 x a B d +12 A a d \right ) \sqrt {\left (b x +a \right )^{2}}\, x}{12 b x +12 a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)*((b*x+a)^2)^(1/2),x)

[Out]

1/12*x*(3*B*b*e*x^3+4*A*b*e*x^2+4*B*a*e*x^2+4*B*b*d*x^2+6*A*a*e*x+6*A*b*d*x+6*B*a*d*x+12*A*a*d)*((b*x+a)^2)^(1
/2)/(b*x+a)

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maxima [B]  time = 0.64, size = 254, normalized size = 1.55 \begin {gather*} \frac {1}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A d x + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a^{2} e x}{2 \, b^{2}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A a d}{2 \, b} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a^{3} e}{2 \, b^{3}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} {\left (B d + A e\right )} a x}{2 \, b} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B e x}{4 \, b^{2}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} {\left (B d + A e\right )} a^{2}}{2 \, b^{2}} - \frac {5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a e}{12 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} {\left (B d + A e\right )}}{3 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)*((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*d*x + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*a^2*e*x/b^2 + 1/2*sqrt(b^2*x^2 +
2*a*b*x + a^2)*A*a*d/b + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*a^3*e/b^3 - 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*(B
*d + A*e)*a*x/b + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*e*x/b^2 - 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*(B*d + A*e
)*a^2/b^2 - 5/12*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*a*e/b^3 + 1/3*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*(B*d + A*e)/b
^2

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mupad [B]  time = 2.92, size = 223, normalized size = 1.36 \begin {gather*} \frac {B\,d\,\left (8\,b^2\,\left (a^2+b^2\,x^2\right )-12\,a^2\,b^2+4\,a\,b^3\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{24\,b^4}+\frac {B\,e\,x\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{4\,b^2}+\frac {A\,\left (a+b\,x\right )\,\left (3\,b\,d-a\,e+2\,b\,e\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{6\,b^2}-\frac {B\,a^2\,e\,\left (\frac {x}{2}+\frac {a}{2\,b}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{4\,b^2}-\frac {5\,B\,a\,e\,\left (8\,b^2\,\left (a^2+b^2\,x^2\right )-12\,a^2\,b^2+4\,a\,b^3\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{96\,b^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)^2)^(1/2)*(A + B*x)*(d + e*x),x)

[Out]

(B*d*(8*b^2*(a^2 + b^2*x^2) - 12*a^2*b^2 + 4*a*b^3*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(24*b^4) + (B*e*x*(a^2
+ b^2*x^2 + 2*a*b*x)^(3/2))/(4*b^2) + (A*(a + b*x)*(3*b*d - a*e + 2*b*e*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(6
*b^2) - (B*a^2*e*(x/2 + a/(2*b))*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(4*b^2) - (5*B*a*e*(8*b^2*(a^2 + b^2*x^2) -
12*a^2*b^2 + 4*a*b^3*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(96*b^5)

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sympy [A]  time = 0.10, size = 63, normalized size = 0.38 \begin {gather*} A a d x + \frac {B b e x^{4}}{4} + x^{3} \left (\frac {A b e}{3} + \frac {B a e}{3} + \frac {B b d}{3}\right ) + x^{2} \left (\frac {A a e}{2} + \frac {A b d}{2} + \frac {B a d}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)*((b*x+a)**2)**(1/2),x)

[Out]

A*a*d*x + B*b*e*x**4/4 + x**3*(A*b*e/3 + B*a*e/3 + B*b*d/3) + x**2*(A*a*e/2 + A*b*d/2 + B*a*d/2)

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